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Unformatted text preview: the sake of variety let’s solve this using diﬀerentials. The problem could be very well solved by writing
the tangent line approximation as above.
√ d. [2 points] Approximate 1.1 using L(x). Solution: If k and x are as above, then √ 1.1 ≈ L(0.1) = 1.05. !
e [2 edge of cube is found to in the approximation error part (d)?
14. .Thepoints] aWhatwasthe error be 30 cm with a possible from in measurement of 0.1 cm. Estimate the maximum
possible error in computing
(a) the volume of the cube
(b) the surface area of the cube.
Solution: Solution: The error is the actual value minus the approximate value which is
1.05 ≈ −0.00119. √ 1.1 − 1 !
15. For what values of r does the function y = erx satisﬁes the diﬀerential equation y + 5y − 6y = 0?
Solution: !
16. If f (x) = 3 + x + ex , ﬁnd (f −1 ) (4).
Solution: !
17. A bacteria culture contains 200 cells initially and grows at a rate proportional to its sze. After half an hour the
population has increased to 360 cells.
(a) Find the number of bacteria after t hours.
Page 1
(b) Find the bacteria after 4 hours.
(c) Find the rate of growth after 4 hours.
(d) When will the population reach 10,000. Page 1
1 Solution: Page 1 ! 18. At what point on the curve y = [ln(x + 4)]2 is the tangent horizontal?
Solution: !
19. (a) Find an equation of the tangent to the curve y = ex that is parallel to the line x − 4y = 1.
(b) Find an equation of the tangent to the curve y = ex that passes through the origin.
Page 1
1
Solution: 1
1 !
20. Find the equation of the tangent line to the curve y = 3 arccos(x/2) at (1, π ).
Solution: The slope of the line is y evaluated at 1. y = − √ 1 2 · 1/2. Hence plugging 1 for x we get
1−(x/2)
√
√
f (1) = − 3. So the equation of the tangent line is y = − 3(x − 1) + π . 21. A cup of hot chocolate has temperature 80◦ C in a room kept at 20◦ C. After half an hour the hot chocolate
cools to 60◦ C.
(a) Write a model describing the temperature of hot chocolate as a function of time. What is the temperature
after another half hour?
(b) When will the chocolate have cooled to 40◦ C?
Solution: 22. How long will it take an investement to double in value if the interest rate is 6% compounded continuously? Find
!
the rate of change of the investment when t = 1.
Solution: ! Page 2 23. Bismuth210 has a halflife of 5 days. A sample population after 1 day is decaying at a rate of 2 mg/day. Find
the initial mass of the sample. Write a model describing the population of the Bismuth210 sample.
Solution: 1/2 = e5k . Hence k = 0.2 ln(0.5). So B = B (0)e0.2 ln(0.5)t . Since the decay rate at t = 1 is 1 mg/day
we know that dB t=1 = 2. Taking derivative of B with respect to t we get
dt
dB
= B (0)0.2 ln(0.5)e0.2 ln(0.5)t .
dt
Plugging t = 1 we get
2 = B (0)0.2 ln(0.5)e0.2 ln(0.5)
So B (0) = 2
0.2 ln(0.5)e0.2 ln 0.5 24. Find the critical points of the function.
(a) f (x) = x3 + 3x2 − 24x (b) h(p) = p−1
p2 +4 (c) f (θ) = 2 cos θ + sin2 θ
(d) f (x) = x ln x
(e) f (x) = xe2x
1 Solution: !
25. Find the local max/min and the absolute max/min of f on the given interval.
(a) f (x) = x4 − 2x2 + 3, [−2, 3] (b) f (x) = x
x2 +4 , [0, 3] (c) f (x) = x − 2 cos x, [−π, π ] (d) f (x) = x − ln x, [1/2, 2] Solution: !
26. Show that 5 is a critical number of the function g (x) = 2 + (x − 5)3 but g does not have a local extreme value at
5.
Solution:
Page 1
115 / Exam 2 (March 20, 2012) page 7 27. Consider the piecewise linear function f (x) graphed below.
! [16 points] Consider the piecewise linear function f (x) graphed below:
Page 1
!
10 f (x) (4, 6) 8 x (10, −6) !
For each function g (x), ﬁnd the value of g (3):
!
a. [4 points] g (For = sinfunction3 g (x), ﬁnd the value of g (x):
x) each [f (x)]
Solution: (a) g (x) = sin(f (x)3 )
(b) g (x) = f (x2 )
x g (x) = cos(f (x)3 ) · 3f (x)2 · f (x) 2
g (3) = f (2) 3
(c) g (x) = ln(f (x)) +cos(7 ) · 3 · 7 · (−1) = 124.0442. (d) g (x) = f −1 (x) b. [4 points] g (x) = f (x2 )
x Solution:
g (x) = x · f (x2 ) · 2x − f (x2 )
x2 ! ! For each function g(x), ﬁnd the value of g (3):
! !
a. [4 points] g(x) = sin [f (x)]3
Solution:
g (x) = cos(f (x)3 ) · 3f (x)2 · f (x) g (3) = cos(73 ) · 3 · 72 · (−1) = 124.0442. b. [4 points] g(x) = f (x2 )
x Solution: x · f (x2 ) · 2x − f (x2 )
x2
3(−3)6 − (−3)
= −5.667.
g (3) =
9
g (x) = c. [4 points] g(x) = ln(f (x)) + f (2)
Solution:
g (x) = 1
f (x) + 0
f (x) g (3) = −1
1
· (−1) =
.
7
7 d. [4 points] g(x) = f −1 (x)
Solution:
g (x) =
g (3) = !
! 1
f (f −1 (x)) 1
2
=− .
f (6)
3 !...
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 Fall '11
 fisherselin

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